Binary operations

Step 1: Put (a*e=a). Step 2: (a+e-1=a) gives (e=1). Step 3: When a constant is subtracted, the identity shifts accordingly.

Step 1: The identity element is (1). Step 2: Put (7*x=1). Then (7+x-1=1), so (x=-5). Step 3: Do not directly use the usual additive inverse; use the given operation.

Step 1: First find (ab). Here (2\cdot2=4). Step 2: The remainder when (4) is divided by (3) is (1). Step 3: In remainder-based operations, the final answer is the remainder, not the original product.

Step 1: Multiplying by (1) keeps the number unchanged. Step 2: In (a*1), (a\cdot1=a), and its remainder modulo (3) is the same (a) for elements of (A). Step 3: For modulo multiplication, check (1) first as identity.

Step 1: Expanding ((a+1)(b+1)) gives (ab+a+b+1). Step 2: Subtracting (1) gives (ab+a+b). Step 3: Such rewritten forms help in associativity and inverse questions.

Step 1: ((1-a)(1-b)=1-a-b+ab). Step 2: (1-(1-a)(1-b)=a+b-ab). Step 3: Rewriting an operation often makes checking its properties easier.

Step 1: Substitute (a=2) and (b=3). Step 2: (2*3=2+3+2\cdot3=11). Step 3: Do not treat the symbol as ordinary multiplication; use the given definition.

Step 1: (2*4=2+3\cdot4=14). Step 2: (4*2=4+3\cdot2=10). Step 3: If changing the order changes the value, the operation is not commutative.

Step 1: (a+b) can range from (2) to (8). Step 2: (\min(a+b,4)) is always one of (2,3,4), all of which belong to (A). Step 3: For closure, match every possible result with the given set.

Step 1: For the (\max) operation, the identity element is (1). Step 2: We need (2*x=1), but (\max(2,x)) can never be (1). Step 3: Elements larger than the identity often do not have inverses in this type of operation.

Step 1: From (a*e=a), (a+e+4=a), so (e=-4). Step 2: For inverse of (10), put (10*x=-4). Then (10+x+4=-4), so (x=-18). Step 3: Keep the identity value correctly while solving.

Step 1: First find the identity (e). From (a+e+4=a), (e=-4). Step 2: Put (10*x=-4). Then (10+x+4=-4), so (x=-18). Step 3: Finding the identity first prevents mistakes in inverse questions.

Step 1: For identity (e), we need (a*e=a). Step 2: (a+e+4=a) gives (e=-4). Step 3: In an operation of the form (a+b+k), the identity is usually (-k).

Step 1: Solve the bracket first. (2*1=2+1+4=7). Step 2: Now (3*7=3+7+4=14). Step 3: In operation questions, always follow the brackets.

Step 1: First find (3*2=3+2+4=9). Step 2: Now (9*1=9+1+4=14). Step 3: Here both bracketings give the same value because this operation is associative.

Step 1: The identity element is (0). Step 2: Put (3*x=0). Then (3+x+3x=0), so (3+4x=0). Step 3: Hence (x=-\frac{3}{4}); solve the linear equation carefully.

Step 1: The identity is (0). Step 2: Put (3*x=0). Then (3+x-3x=0), so (3-2x=0). Step 3: Hence (x=\frac{3}{2}); be careful with signs.

Step 1: For (a*e=a), we need (ae=a). Step 2: Since (a>0), (e=1). Step 3: In ordinary multiplication, the identity element is (1).

Step 1: The identity for multiplication is (1). Step 2: (5*x=1) gives (5x=1), so (x=\frac{1}{5}). Step 3: For multiplicative inverse, take the reciprocal.

Step 1: The identity for multiplication is (1). Step 2: (2*x=1) gives (x=\frac{1}{2}), which is not an integer. Step 3: The inverse must belong to the same set.

Step 1: The identity element is (0). Step 2: From (a*x=0), (a+x+ax=0), so (x=-\frac{a}{1+a}). Step 3: When (a=-1), the denominator becomes (0), so the inverse does not exist.

Step 1: The identity element is (0). Step 2: From (a*x=0), (a+x-ax=0), so (x=-\frac{a}{1-a}). Step 3: When (a=1), the denominator becomes (0), so the inverse does not exist.

Step 1: For identity (e), we need (\gcd(a,e)=a). Step 2: This happens when (e) is a multiple of every (a), and (8) works in the set. Step 3: For a (\gcd) operation on divisors of a number, the identity is often the largest element.

Step 1: For identity (e), we need (\operatorname{lcm}(a,e)=a). Step 2: (\operatorname{lcm}(a,1)=a) for every (a \in A). Step 3: For an (\operatorname{lcm}) operation, checking (1) first is useful.

Step 1: (1*2=2\cdot1+2=4), so ((1*2)*3=4*3=11). Step 2: (2*3=7), so (1*(2*3)=1*7=9). Step 3: If changing brackets changes the value, the operation is not associative.